Compare two LED driver

Discussion in 'General Electronics Chat' started by roboticvn, Aug 4, 2014.

  1. roboticvn

    Thread Starter Member

    Jul 9, 2014
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    I want to make a LED driver for a while LED (1W, 350mA, forward voltage: 3.1-3.4V). I find two circuit on the Internet. However, I don't know advantage and disadvantage of them. I hope you talk for me should should choose which. Thank you very much^^
    Firsly, the circuit using IC LM317
    [​IMG]
    Other one,
    [​IMG]
     
  2. to3metalcan

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    Jul 20, 2014
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    The only "disadvantage" of the top circuit is the slightly-more-expensive LM317 regulator. It's probably the more high-performance of the two, strictly speaking.

    The bottom circuit uses cheaper parts, but it's also a bit more complicated.

    They both appear to do about the same thing (drive a load at a regulated current.)

    Pretty sure you could use the IRF510 for either, which is maybe an advantage if you're in the US, since you can buy 'em at RadioShack.
     
  3. roboticvn

    Thread Starter Member

    Jul 9, 2014
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    Thank for your help^^
     
  4. roboticvn

    Thread Starter Member

    Jul 9, 2014
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    Can you explain for me how does Q3 activity? And compare power dissipation of two circuit? Thank you
     
  5. to3metalcan

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    Jul 20, 2014
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    Q3 looks like just a buffer stage so that a) your signal is conducted to ground instead of hot with a minimum number of components, and b) the current-limiting action of Q2 doesn't mess with the impedance your PWM signal sees.
     
  6. roboticvn

    Thread Starter Member

    Jul 9, 2014
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    Can you explain more clearly? When does Q3 on and when does Q3 off? Thank you. I read on Internet that when provice ~0.7V to Q3 through R3, Q3 off but I don't understand.
     
  7. Sensacell

    Well-Known Member

    Jun 19, 2012
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    The FET is biased on by R1, when the current reaches a level that produces about 0.7 V across R2, Q2 will start to turn on, which effectively steals the bias from the FET. This circuit will have relatively poor initial predictability, it will also be sensitive to temperature changes.

    Q3 acts as an alternate, external way to turn the FET off, in the same way that Q2 does.

    The LM317 has an internal voltage reference, it will provide a more accurate and stable current.
     
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  8. to3metalcan

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    Q3 is biased on, albeit somewhat unpredictably (the device's beta is going to set the operating point with a biasing setup this minimal.) I would not expect it to be turned off by .7V.
     
  9. ScottWang

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    Aug 23, 2012
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    If you want to using the bottom circuit then you should change the values of resistors, for a 12 Vcc power, the R1 could using 10K and the R3 can using 1K, if the Re(R1) too large then the Ib will be too low, it could cause the Ic not enough, and the saturation status will be in an astable status.

    The Q2 was designed to turn off the Q1 when the Id is too high via the R2 to detect Id, if you want to using this circuit for your LED, then you should in series a limiting resistor with S of Q1.

    If you thinking in logical way, when the input is a low level, the Q3 will be turn on, and the Ve of Q3 is close to 0.9V, this Vgs voltage less then 2V, the Vg is similar connecting to a ground, the Vgs <2V will cause the Q1 to turn off, when the input is a high level, the Q3 will be turn off, and the Q1 will be turn on, but the problem here is the E of Q3 is connecting to a voltage higher then the input voltage, so this could cause the Q3 may not turn off.
     
  10. to3metalcan

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    Jul 20, 2014
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    I agree, R1 seems big, but I'm not sure Q2 can limit the gate voltage the way it's supposed to if you tamper with its value too much. You might have to raise the value of R2 to compensate for the change in response. Also, I don't think you need a series resistor for this circuit (though it might not hurt anything.) The idea is that Q2 limits the maximum current Q1 can deliver per pulse.
     
  11. ScottWang

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    Aug 23, 2012
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    The Q2 turn on is depends on the current flows through the R2, no matter with R1, the R1 just a pull high for the mosfet providing a positive for Vg to turn on.

    The R2 is to provide the Vbe for Q2, and that's the Imax, if the Imax cause Q2_Vbe=0.7V, the Voltage(Vcc-0.7V-0.0945V) cross on the LEDs could be too high, it might destroy the LEDs, those are depend on the Vcc and the voltage range of LEDs(Vf), that's why i want to in series with a resistor to limiting the LED current, if the circuit for a relay maybe ok, because the relay has a big range of voltage.

    The Vds = 350 mA * 0.6Ω = 0.0945V
     
    Last edited: Aug 5, 2014
  12. roboticvn

    Thread Starter Member

    Jul 9, 2014
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    Thank you for your help but I'm sorry, I still don't understand. When input is low (0V), Vbase(Q3)=0; Vcollector(Q3)=0 (ground), Ve(now)=? Why does it conduct? I know that normal PNP transistor start on when Veb~0.7V. Is that true? I hope you can help me. Thank you
     
  13. to3metalcan

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    Yes, that's true. The emitter voltage is going to be high, relative to the base.
     
  14. to3metalcan

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    Scott - R1 does matter, though...it's the top half of the voltage divider Q2 is the lower half of. Shrinking it is going to raise the gate voltage, which means Q2 has to conduct more to bring it down...raising the value of R2 will compensate for this by requiring less current to turn on Q2.

    You're right, though...without knowing the supply voltage, speculation is idle. Robotic, what voltage are you running the circuit at?
     
  15. roboticvn

    Thread Starter Member

    Jul 9, 2014
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    Thank for your help^^ This is a circuit I refer on the Internet. I use R2=1.6R, 5V supply, I simulated on Proteus and creat 325mA current
     
  16. ScottWang

    Moderator

    Aug 23, 2012
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    Do some calculation:
    Q3_Ib = 9V - Q3_Vbe / 1M + 4.7K = (9V - 0.7V)/1004.7K = 0.000008261A = 8.26uA
    V_R3 = 8.26uA * 4.7K = 0.0388V
    V_R1 = 8.26uA * 1M = 8.26uA * 1000000Ω = 8.26V
    Because the Ib is too low, that's why i was said that you need to adjust the values of resistors.

    You can see the below showing the current and voltages.
    [​IMG]
     
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  17. Sensacell

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    Jun 19, 2012
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    It occurs to me is that things would be much simpler if Q3 was replaced with a NPN transistor with the emitter grounded.

    Otherwise you need to drive the input with a voltage almost equal to the supply to activate the current sink- not Arduino compatible
     
  18. to3metalcan

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    I think the OP indicated a 5V supply?
     
  19. Sensacell

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    If that is the case, the IRF520 will not get enough Vgs to fully turn on.
     
  20. ScottWang

    Moderator

    Aug 23, 2012
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    The circuit was working as logical voltage, only the low and high, and the Q2 as a switch not a resistor to do the voltage divider.

    The circuit was used for the current source component, if the power is providing a voltage source, it could be damage the LEDs, you can see the circuit and tell me how do you decide the +V power and the Vf of LEDs?

    [​IMG]
     
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